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Is total mechanical energy, i.e. Kinetic Energy + Potential Energy, conserved in a frame which is moving with constant velocity with respect to earth.

Consider a ball dropped from a building. The ball and earth are the system.

Let us consider two frames. One attached to the earth and the other moving with respect to the earth at constant velocity, say 1 m/s. For a time interval.

If total mechanical energy is individually conserved in the two frames, then that means that the loss in potential energy in the respective frames is equal to the magnitude of change in kinetic energy in the same frame. Now since change in kinetic energy is different in both, the change in potential energy must also be different. But as I have seen in numerous answers and texts, potential energy depends upon the configuration of particles of the system, so it is frame independent.

This seems to imply that total mechanical energy is not conserved in all inertial frames.

Then again, if we define the potential energy as the work done by gravitational force, since that is different in different frames, due to distance through which force acts in different frames....then mechanical energy is conserved and the equations are satisfied. But then that would mean that the potential energy is defined with respect to the frame.

So, finally, what is true and where am I wrong?

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  • $\begingroup$ is your frame moving in perpendicular or parallel direction to the direction of fall of the ball ?? $\endgroup$
    – Ankit
    Aug 27, 2020 at 12:11
  • $\begingroup$ Sorry didn't mention... parallel. $\endgroup$ Aug 27, 2020 at 12:24
  • $\begingroup$ Was the frame moving from earlier or both started moving simultaneously ? $\endgroup$
    – Ankit
    Aug 27, 2020 at 12:29
  • $\begingroup$ Yes...but work done by gravitational force does not equal mg∆h in this case $\endgroup$ Aug 27, 2020 at 14:47
  • $\begingroup$ I think this answer addresses your concerns since the evaluation of energy is independent of the location where it is evaluated. Energy is an invariant quantity. $\endgroup$ Aug 28, 2020 at 13:13

3 Answers 3

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TLDR: In Newtonian mechanics the change in potential energy is frame independent and, in the absence of dissipation, total mechanical energy is conserved in all frames. The confusion you are encountering is not due to any problem with the definition of potential energy or mechanical energy, it is due to not considering the total mechanical energy.

Unfortunately, I must disagree with the other answer. This is a confusing topic, but it is incorrect that:

the work done by gravity (which is now also the change in potential energy)

The change in potential energy remains $mg\Delta h$ (where $h$ is the separation) although the work done by gravity does not. The potential energy must remain the same or you could create a perpetual motion machine by for example compressing a spring at rest and then releasing it while moving.

Let's investigate this more fully. For small changes in separation, $\Delta h$ between the ball and the earth the change in gravitational potential energy is $\Delta U = mg\Delta h$. Note that this potential energy does not belong to the ball itself, but rather to the ball-earth system. In other words, $\Delta h$ is not a property of the ball only, but a property of the ball-earth system, the configuration of the two. That means that if you are considering the gravitational potential energy then you must consider the entire system of both the ball and the earth. The other analyses omitted the earth.

The mechanical energy is the sum of the kinetic energy of the Earth, the kinetic energy of the ball, and the potential energy, $E=T_E+T_B+U$, where $T_E=\frac{1}{2}M V^2$, $T_B=\frac{1}{2} m v^2$, and $U=mgh$. In a frame where the earth and ball are initially moving at $u$ it can be shown that $$T_E(t)=\frac{1}{2}Mu^2+mgut+\frac{1}{2}\frac{m^2}{M}g^2 t^2$$ $$T_B(t)=\frac{1}{2}mu^2-mgut+\frac{1}{2}mg^2 t^2$$ $$U(t)=mgh_0 - \frac{1}{2}\left(\frac{m^2}{M}+m\right)g^2 t^2$$ $$E(t)=\frac{1}{2}(m+M)u^2+mgh_0=const.$$ So the total mechanical energy $E=T_E+T_B+U$ is conserved, but $T_B+U$ is not.

Given this, why is it that we can get away with using only $T_B+U$ in the frame where $u=0$? Note, if $m\ll M$ then the last term of $T_E$ drops out and it simplifies to $$T_E=\frac{1}{2}Mu^2+mgut$$ This is non-zero except in the case where $u=0$ and more importantly it changes over time if $u\ne 0$. So we can get away with the "cheat" because the Earth is much more massive than the ball and so that leads to $T_E=0=const.$ but only in the frame where $u=0$.

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Normally we say that the change in potential energy of an object is $mg$ times the change in height, and we equate this change in potential energy with the work done by gravity on that object. This is true in a frame of reference which is stationary (or, at least, has zero vertical velocity) with respect to the earth - which is why it is very convenient to work in such a frame of reference, and we almost always do so.

However, in a frame of reference which has a non-zero vertical velocity with respect to the earth, $mg \Delta h$ is no longer equal to the work done by gravity, and we must go back to first principles.

For example, if an object with mass $m$ has initial velocity $u$ (upwards) with respect to the earth, then we could work in a frame of reference in which the object is initially stationary. We know that in this frame of reference the object's velocity at time $t$ is $-gt$, so its change in kinetic energy is $ \frac 1 2 m g^2 t^2$. Its displacement at time $t$ is $-\frac 1 2 g t^2$, so the work done by gravity on the object is

$(-mg) \times (- \frac 1 2 g t^2) = \frac 1 2 m g^2 t^2$

and so we see that the work done by gravity is equal to the change in kinetic energy, as we expect.

Alternatively, we could work in a frame of reference that is stationary with respect to the earth. In this frame the object has an initial kinetic energy of $\frac 1 2 mu^2$ and a final kinetic energy of $\frac 1 2 m (u-gt)^2$, so the change in kinetic energy is

$\frac 1 2 m \left( (u-gt)^2 - u^2 \right) = \frac 1 2 mg^2 t^2 - mugt$

In this frame the displacement of the object at time $t$ is $ut - \frac 1 2 gt^2$ so the work done by gravity (which is now also the change in potential energy) is

$(-mg) \times (ut - \frac 1 2 gt^2) = \frac 1 2 mg^2t^2 - mugt$

and, once again, we see that the work done by gravity is equal to the change in kinetic energy.

So the principle that the work done by gravity is equal to the change in kinetic energy (in the absence of friction or other dissipative forces) is true in both frames, although the value on each side of the equation is frame dependent.

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  • $\begingroup$ Yeah that I was able to calculate. The doubt is regarding definitions of terms. So mechanical energy is not conserved...since you say change in potential energy is not equal the work done by gravity? Since mechanical energy is still PE + KE, right? $\endgroup$ Aug 27, 2020 at 10:57
  • $\begingroup$ @AbsoluteZero It is true that $mgh$ + KE is not conserved. You can still say that PE + KE is conserved as long as you redefine PE to be $mg \Delta d$ where $\Delta d$ is displacement measured relative to the frame of reference instead of relative to the earth. $\endgroup$
    – gandalf61
    Aug 27, 2020 at 13:50
  • $\begingroup$ This all seems semantic and not of very much use now... $\endgroup$ Aug 27, 2020 at 14:51
  • $\begingroup$ @AbsoluteZero I tend to agree. Understanding the underlying principles of work and energy is more important than the terminology. $\endgroup$
    – gandalf61
    Aug 27, 2020 at 15:07
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In a paper, I have shown that, "in general, in an inertial frame of reference, in the presence of nonconservative forces such as frictional, normal, and tension forces, if the work done by nonconservative forces on a system is zero, the mechanical energy is conserved. The work done by the same nonconservative forces in another inertial reference frame may be nonzero. Therefore, in the second frame of reference mechanical energy is not conserved".

The nonconservative work-mechanical energy theorem defined by $∆E = W_{nc}$ is covariant under the Galilean transformations.

The mechanical energy of a system can be conserved in a given inertial frame of reference, and not be conserved in another inertial reference frame. In the absence of the nonconservative forces, the mechanical energy is conserved in all inertial reference frames.

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